Question

What is the are of point ABCD if there coordinats are (A, 1 1/2, 1) (B, 1 1/2, 6 1/3) (C, 5 1/2, 6 1/3) (D, 5 1/2, 1)

Answer 1

The answer is below

Explanation:

What is the area of ABCD if the coordinates of the vertices are A (1 1/2, 1) = (3/2, 1), B(1 1/2, 6 1/3) = (3/2, 19/3), C(5 1/2, 6 1/3) = (11/2, 19/3) and D = (5 1/2, 1) = (11/2, 1). Hence A(3/2, 1), B(3/2, 19/3), C(11/2, 19/3) and D(11/2, 1)

Therefore from the vertices, we can see that ABCD is a rectangle.

Area of ABCD = AB * BC

The distance between two points
`(x_1,y_1)\ a nd\ (x_2,y_2)` in the coordinate plane is given by:

`Distance=√((x_2-x_1)^2+(y_2-y_1)^2)`

Therefore:

`AB=\sqrt{((3)/(2)-(3)/(2) )^2+((19)/(3)-1)^2} =(16)/(3) \ units\\\\BC=\sqrt{((11)/(2)-(3)/(2) )^2+((19)/(3)-(19)/(2) )^2} =4\ units`

Area of ABCD = AB * BC = 16/3 units * 4 units = 64/3 unit²